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KRYPTON GROUND STATE ENERGIES
By Prof. L. Kaliambos (Natural Philosopher in New Energy) December 21, 2015 Krypton is an atom of the chemical element krypton with symbol Kr and atomic number 36. However unlike for hydrogen atom, a closed-form solution to the Schrödinger equation for the many-electron atoms has not been found. So, under the invalid relativity ( EXPERIMENTS REJECT RELATIVITY) various approximations, such as the Hartree–Fock method, could be used to estimate the ground state energies. Under these difficulties I analyzed carefully the electromagnetic interactions of two spinning electrons of opposite spin which give a simple formula for the solution of such ground state energies. You can see it in my published paper “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures”(2008). Under this condition today it is well known that the correct electron configuration should be given by this correct image of Krypton including the following electron configuration: 1s2.2s2.2px2.2py2.2pz2.3s2.3px2.3py2.3pz2.3d10. 4s2. 4p6 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of krypton (from (E1 to E26) are the following: E1 = 13.99, E2 = 24.36, E3 = 36.95 , E4 = 52.5 , E5 = 64.7, E6 = 78.5, E7 = 111, E8= 125.8, E9 = 230.85, E10 = 268.2, E11 = 308 , E12 = 350, E13 = 391, E14 = 447 , E15 = 492, E16 = 541, E17 = 592, E18 = 641, E19 = 786, E20 = 833 , E21 = 884 E22= 937, E23 = 998 , E24 = 1051, E25 = 1151, E26 = 1205.3 Firstly we examine the -( E1 + E2 + Ε3 + Ε4 + Ε5 + Ε6 ) = -271 = E(4p6) Here the E(4p6) represents the ground state energy of the six outermost electrons. Then, we observe that -( Ε7 + E8 ) = - 236.8 = E(4s2) - ( E9 + … + E18 ) = - 4261 = E(3d10). -(E19 +…+ E24) = - 5489 = E(3p6) -(E25 + E26 ) = -2356.3 = E(3s2) It is of interest to note that in the absence of data (from E27 to E36) one cannot calculate the following theoretical ground state energies : -( E27 +…+ E32 ) = E(2p6) -(E33 + E34) = E(2s2) -(E35 + E36) = E(1s2) For understanding better the ground state energies see also my papers about the ground state energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. ' ' GROUND STATE ENERGY OF - ( E1 +…+ E6 ) = -271 = E(4p6) Here the ground state energy E(4p6) of the six outermost electrons is given by applying my formula of 2008. The charges (-30e) of the electrons (1s22s22p63s23p63d104s2) screen the nuclear charge (+36e) and for a perfect screening we would have an effective ζ = 6. However the electrons 4p6 of the three orbitals repel the 4s2 leading to the deformation of 4s2 . Thus ζ > 6. Under this condition we may write (E1 +… + E6) = 271 eV = -E(4p6) = -327.21 )ζ2 + (16.95) ζ - 4.1 / n2 Since n = 4 the above equation could be written as 5.1 ζ2 – 3.178 ζ - 270.23 = 0 Then solving for ζ we get ζ = 7.6 > 6 ' ' GROUND STATE ENERGY OF -( Ε7 + E8 ) = - 236.8 = E(4s2) ''' Here the E(4s2) represents the ground state energy of the two electrons with opposite spin given by applying my formula of 2008. The charges (-28e) of the inner electrons (1s22s22p63s23p63d10) screen the nuclear charge (+36e) and for a perfect screening we would have ζ = 8. However according to the quantum mechanics the two electrons of 4s2penetrate the 3d10 and lead to the deformation of 3d10 . Thus we must observe that ζ > 8. Under this condition we may write (Ε7 + E8 ) = 236.8 = -E(4s2) = - + (16.95)ζ - 4.1 / n2 Since n = 4 we may rewrite 1.7ζ2 -1.059ζ -236.54 = 0 Then solving for ζ we get ζ = 12.11 > 8 ' ' '''GROUND STATE ENERGY OF ( E9 + … + E18 ) = 4261 eV = -E(3d10) Here the E(3d10) represents the ground state energy of the 10 electrons (3d10) . Note that the 10 electrons create five orbitals with electrons of opposite spin given by applying my formula of 2008. The charges (-18e) of the inner electrons (1s22s22p63s23p6) screen the nuclear charge (+36e) and for a perfect screening we would have ζ = 18. Under this condition we may write ' '''E9 +..+ E18 ) = 4261 eV = - E(3d10) = -5+ (16.95)ζ - 4.1 / n2 So using n = 3 we may rewrite 15.1167ζ2- 9.4167ζ - 4258.72 = 0 Surprisingly solving for ζ we get ζ = 17.01 < 18 , which cannot exist . In fact, the 10 electrons of the five orbitals make a complete spherical sub- shell leading to a perfect screening with ζ = 18. Thus using ζ = 18 we expect to determine n > 3. Under this condition we may write E9 +..+ E18 ) = 4261 eV = -E(3d10) = -527.21)182 - (16.95)18 + (4.1) / n2 Then solving for n we get n = 3.16 > 3 . ' ' '''GROUND STATE ENERGY OF ( E19 + … + E24 ) = 5489 eV = -E(3p6)' Here the E(3p6) represents the ground state energy of the 6 paired electrons given by applying my formula of 2008. ' '''The charges (-12e) of the twelve inner electrons like (1s22s22p63s2 ) screen the nuclear charge (+36e) and for a perfect screening we would have an effective Zeff = ζ = 24. However the 6 paired electrons ( 3p6 ) repel the 3s2 electrons and lead to the deformation of shells with ζ > 24. Under this condition we may write ( E19 +…+ E24) = 5489 eV' = '-E(3p6) '= '-3+(16.95)ζ - 4.1 / n2 Since n = 3 the above equation can be written as 9.07ζ2 - 5.65ζ - 5487.63 = 0 Then, solving for ζ we get ζ = 24.91 > 24 . ' ' '''GROUND STATE ENERGY OF ( E25 + E26 ) = 2356.3 eV = -E(3s2)' Here the E(3s2) represents the ground state energy of the two paired electrons (3s2) given by applying my formula of 2008. The charges (-10e) of the inner 10 electrons of 1s2.2s2.2p6 screen the nuclear charge (+36e) and for a perfect screening we would have an effective ζ = 26. However the two electrons of 3s2 penetrate the 2p6 leading to the deformation of shells with ζ > 26. Under this condition we write the following equation as ( E25 + E26 ) = 2356.3 eV = - E(3s2) = - 27.21)ζ2 + (16.95) ζ - 4.1 / n2 Since n = 3 we may write 3.0233ζ2 - 1.8833ζ - 2355.84 = 0 Then solving for ζ we get ζ = 28.23 > 26 . Note that the two electrons of opposite spin (4s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. However in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. 'WHY EINSTEIN’S RELATIVITY LED TO THE ABANDONMENT OF LAWS IN FAVOR OF WRONG ATOMIC THEORIES ' It is of interest to note that in the absence of a detailed knowledge about the electromagnetic interaction of two electrons of opposite spin physicists so far under the influence of Einstein’s relativity could not calculate such ground state energies of many-electron atoms. Historically, despite the enormous success of the Bohr model and the quantum mechanics of the Schrodinger equation based on the well-established laws of electromagnetism in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far, under the influence of Einstein’s relativity which led to the abandonment of natural laws neither was able to provide a satisfactory explanation of the two-electron atoms. In atomic physics a two-electron atom is a quantum mechanical system consisting of one nucleus with a charge Ze and just two electrons. This is the first case of many-electron systems. The first few two-electron atoms are: Z =1 : H- hydrogen anion. Z = 2 : He helium atom. Z = 3 : Li+ lithium atom anion. Z = 4 : Be2+ beryllium ion. Prior to the development of quantum mechanics, an atom with many electrons was portrayed like the solar system, with the electrons representing the planets circulating about the nuclear “sun”. In the solar system, the gravitational interaction between planets is quite small compared with that between any planet and the very massive sun; interplanetary interactions can, therefore, be treated as small perturbations. However, In the helium atom with two electrons, the interaction energy between the two spinning electrons and between an electron and the nucleus are almost of the same magnitude, and a perturbation approach is inapplicable. In 1925 the two young Dutch physicists Uhlenbeck and Goudsmit discovered the electron spin according to which the peripheral velocity of a spinning electron is greater than the speed of light. Since this discovery invalidates Einstein’s relativity it met much opposition by physicists including Pauli. Under the influence of Einstein’s invalid relativity physicists believed that in nature cannot exist velocities faster than the speed of light.(See my FASTER THAN LIGHT). So, great physicists like Pauli, Heisenberg, and Dirac abandoned the natural laws of electromagnetism in favor of wrong theories including qualitative approaches under an idea of symmetry properties between the two electrons of opposite spin which lead to many complications. Thus, in the “Helium atom-Wikipedia” one reads: “Unlike for hydrogen a closed form solution to the Schrodinger equation for the helium atom has not been found. However various approximations such as the Hartree-Fock method ,can be used to estimate the ground state energy and wave function of atoms”. It is of interest to note that in 1993 in Olympia of Greece I presented at the international conference “Frontiers of fundamental physics” my paper “Impact of Maxwell’s equation of displacement current on electromagnetic laws and comparison of the Maxwellian waves with our model of dipolic particles ". In that paper I showed that LAWS AND EXPERIMENTS INVALIDATE FIELDS AND RELATIVITY . At the same period I tried to find not only the nuclear force and structure but also the coupling of two electrons under the application of the abandoned electromagnetic laws. For example in the photoelectric effect the absorption of light contributed not only to the increase of the electron energy but also to the increase of the electron mass, because the particles of light have mass m = hν/c2. ( See my paper "DISCOVERY OF PHOTON MASS" ). However the electron spin which gives a peripheral velocity greater than the speed of light cannot be affected by the photon absorption. Thus after 10 years I published my paper "Nuclear structure..electromagnetism" (2003), in which I showed not only my DISCOVERY OF NUCLEAR FORCE AND STRUCTURE but also that the peripheral velocity (u >> c) of two spinning electrons with opposite spin gives an attractive magnetic force (Fm) stronger than the electric repulsion (Fe) when the two electrons of mass m and charge (-e) are at a very short separation (r < 578.8 /1015 m). Because of the antiparallel spin along the radial direction the interaction of the electron charges gives an electromagnetic force Fem = Fe - Fm . Therefore in my research the integration for calculating the mutual Fem led to the following relation: Fem = Fe - Fm = Ke2/r2 - (Ke2/r4)(9h2/16π2m2c2) Of course for Fe = Fm one gets the equilibrium separation ro = 3h/4πmc = 578.8/1015 m. That is, for an interelectron separation r < 578.8/1015 m the two electrons of opposite spin exert an attractive electromagnetic force, because the attractive Fm is stronger than the repulsive Fe . Here Fm is a spin-dependent force of short range. As a consequence this situation provides the physical basis for understanding the pairing of two electrons described qualitatively by the Pauli principle, which cannot be applied in the simplest case of the deuteron in nuclear physics, because the binding energy between the two spinning nucleons occurs when the spin is not opposite (S=0) but parallel (S=1). According to the experiments in the case of two electrons with antiparallel spin the presence of a very strong external magnetic field gives parallel spin (S=1) with electric and magnetic repulsions given by Fem = Fe + Fm So, according to the well-established laws of electromagnetism after a detailed analysis of paired electrons in two-electron atoms I concluded that at r < 578.8/1015 m a motional EMF produces vibrations of paired electrons. Unfortunately today many physicists in the absence of a detailed knowledge believe that the two electrons of two-electron atoms under the Coulomb repulsion between the electrons move not together as one particle but as separated particles possessing the two opposite points of the diameter of the orbit around the nucleus. In fact, the two electrons of opposite spin behave like one particle circulating about the nucleus under the rules of quantum mechanics forming two-electron orbitals in helium, beryllium etc. In my paper of 2008, I showed that the positive vibration energy (Ev) described in eV depends on the Ze charge of nucleus as Ev = 16.95Z - 4.1 Of course in the absence of such a vibration energy Ev it is well-known that the ground state energy E described in eV for two orbiting electrons could be given by the Bohr model as E = (-27.21) Z2. So the combination of the energies of the Bohr model and the vibration energies due to the opposite spin of two electrons led to my discovery of the ground state energy of two-electron atoms given by -E = (-27.21) Z2 + (16.95 )Z - 4.1 For example the laboratory measurement of the ionization energy of H- yields an energy of the ground state -E = - 14.35 eV. In this case since Z = 1 we get -E = -27.21 + 16.95 - 4.1 = -14.35 eV. In the same way writing for the helium Z = 2 we get -E = - 108.8 + 32.9 - 4.1 = -79.0 eV The discovery of this simple formula based on the well-established laws of electromagnetism was the first fundamental equation for understanding the energies of many-electron atoms, while various theories based on qualitative symmetry properties lead to complications. Category:Fundamental physics concepts